Question: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $r \neq 0$. $x = \dfrac{-5r + 45}{3r - 6} \times \dfrac{7r - 7}{r^2 - 10r + 9} $
Answer: First factor the quadratic. $x = \dfrac{-5r + 45}{3r - 6} \times \dfrac{7r - 7}{(r - 9)(r - 1)} $ Then factor out any other terms. $x = \dfrac{-5(r - 9)}{3(r - 2)} \times \dfrac{7(r - 1)}{(r - 9)(r - 1)} $ Then multiply the two numerators and multiply the two denominators. $x = \dfrac{ -5(r - 9) \times 7(r - 1) } { 3(r - 2) \times (r - 9)(r - 1) } $ $x = \dfrac{ -35(r - 9)(r - 1)}{ 3(r - 2)(r - 9)(r - 1)} $ Notice that $(r - 1)$ and $(r - 9)$ appear in both the numerator and denominator so we can cancel them. $x = \dfrac{ -35\cancel{(r - 9)}(r - 1)}{ 3(r - 2)\cancel{(r - 9)}(r - 1)} $ We are dividing by $r - 9$ , so $r - 9 \neq 0$ Therefore, $r \neq 9$ $x = \dfrac{ -35\cancel{(r - 9)}\cancel{(r - 1)}}{ 3(r - 2)\cancel{(r - 9)}\cancel{(r - 1)}} $ We are dividing by $r - 1$ , so $r - 1 \neq 0$ Therefore, $r \neq 1$ $x = \dfrac{-35}{3(r - 2)} ; \space r \neq 9 ; \space r \neq 1 $